What is area x=2y,x+1=(y-1)^2 ?

The answer to area x=2y,x+1=(y-1)^2 is 32/3

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area x=2y,x+1=(y-1)^2

area

x = 2 y, x + 1 = (y - 1)^{2}

\frac{32}{3}

Decimal

10.66666 \dots

Solution steps

Isolate

y

for

x = 2 y : y = \frac{x}{2}

Isolate

y

for

x + 1 = (y - 1)^{2} : y = x + 1 + 1, y = - x + 1 + 1

f_{1} (x) = \frac{x}{2}

f_{2} (x) = x + 1 + 1

To find the intersection points solve

f_{i} (x) = f_{j} (x)

\frac{x}{2} = x + 1 + 1 : x = 8

\frac{x}{2} = - x + 1 + 1 : x = 0

x + 1 + 1 = - x + 1 + 1 : x = - 1

= \int_{- 1}^{0} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{0}^{8} ∣ f_{1} (x) - f_{2} (x) ∣ d x

= \int_{- 1}^{0} x + 1 + 1 - (- x + 1 + 1) d x + \int_{0}^{8} \frac{x}{2} - (x + 1 + 1) d x

\int_{- 1}^{0} x + 1 + 1 - (- x + 1 + 1) d x = \frac{4}{3}

\int_{0}^{8} \frac{x}{2} - (x + 1 + 1) d x = \frac{28}{3}

= \frac{4}{3} + \frac{28}{3}

Simplify

= \frac{32}{3}

5 - x^{2}, x^{2} - 3

\int_{0}^{4} ∣ 2 x - 6 ∣ d x

\frac{\partial}{\partial x} (ln (x y) + ln (\frac{x}{y}))

\int (e^{2 y}) d y

\frac{d}{d x} (7^{x^{4} + 2 x})